29. Half Adder using NAND Primitive

Implement a 1-bit half adder structurally using only the nand primitive. The outputs are sum = a ⊕ b and cout = a & b.

Requirements

Module name: ha_nand_only
Inputs: a, b
Outputs: sum, cout
Rules: use only nand primitives; purely structural; no procedural blocks

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Testbench Code

`timescale 1ns/1ps
module tb_ha_nand_only;
  reg  a, b;
  wire sum, cout;

  wire expected_sum, expected_cout;
  wire mismatch;

  ha_nand_only dut(.a(a), .b(b), .sum(sum), .cout(cout));

  assign expected_sum  = a ^ b;
  assign expected_cout = a & b;

  assign mismatch = (sum !== expected_sum) | (cout !== expected_cout);

  initial begin
    $dumpfile("tb_ha_nand_only.vcd");
    $dumpvars(0,
      a, b,
      sum, cout,
      expected_sum, expected_cout,
      mismatch
    );
  end

  integer TOTAL_TEST_CASES, TOTAL_PASSED_TEST_CASES, TOTAL_FAILED_TEST_CASES;
  integer ia, ib;

  initial begin
    TOTAL_TEST_CASES = 0; TOTAL_PASSED_TEST_CASES = 0; TOTAL_FAILED_TEST_CASES = 0;

    for (ia = 0; ia < 2; ia = ia + 1) begin
      for (ib = 0; ib < 2; ib = ib + 1) begin
        a = ia[0]; b = ib[0]; #1;
        $display("CASE=a%0d_b%0d : sum=%0b expected_sum=%0b  cout=%0b expected_cout=%0b  %s",
                 a, b, sum, expected_sum, cout, expected_cout, mismatch ? "MISMATCH" : "OK");
        TOTAL_TEST_CASES = TOTAL_TEST_CASES + 1;
        if (!mismatch) TOTAL_PASSED_TEST_CASES = TOTAL_PASSED_TEST_CASES + 1;
        else begin
          TOTAL_FAILED_TEST_CASES = TOTAL_FAILED_TEST_CASES + 1;
          $display("  FAILED INPUTS: a=%0b b=%0b", a, b);
        end
        #4;
      end
    end

    $display("TOTAL_TEST_CASES=%0d",        TOTAL_TEST_CASES);
    $display("TOTAL_PASSED_TEST_CASES=%0d", TOTAL_PASSED_TEST_CASES);
    $display("TOTAL_FAILED_TEST_CASES=%0d", TOTAL_FAILED_TEST_CASES);
    $display("ALL_TEST_CASES_PASSED=%s",    (TOTAL_FAILED_TEST_CASES==0) ? "true" : "false");
    $finish;
  end
endmodule